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2 May, 19:16

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want to know how many seconds the basketball is in the air before it hits the ground. We can ignore air resistance.

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Answers (2)
  1. 2 May, 20:35
    0
    t = 2.32 s

    Explanation:

    Applying the equation of motion;

    d = ut + 0.5gt^2

    Where;

    d = distance travelled

    u = initial velocity

    g = acceleration due to gravity

    t = time taken

    Since the object was dropped;

    u = 0

    Then,

    d = 0.5gt^2

    t^2 = d/0.5g

    t = √ (d/0.5g) ... 1

    Given

    g = 3.7 m/s^2

    d = 10 m

    Substituting the values;

    t = √ (10 / (0.5*3.7))

    t = 2.32 s
  2. 2 May, 21:42
    0
    2.32 s

    Explanation:

    Using the equation of motion,

    s = ut+g't²/2 ... Equation 1

    Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.

    Note: Since Onur drops the basket ball from a height, u = 0 m/s

    Then,

    s = g't²/2

    make t the subject of the equation,

    t = √ (2s/g') ... Equation 2

    Given: s = 10 m, g' = 3.7 m/s²

    Substitute this value into equation 2

    t = √ (2*10/3.7)

    t = √ (20/3.7)

    t = √ (5.405)

    t = 2.32 s.
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